How do you find the average rate of change of #g(x)=1/(x-2)# over #[x, x+h]# ?
1 Answer
Explanation:
The average rate of change is the slope of the secant between:
#(x, g(x))# and#(x+h, g(x+h))#
The slope
#m = (y_2-y_1)/(x_2-x_1)#
So the slope of the secant is:
#(g(x+h)-g(x))/((x+h)-x) = (g(x+h)-g(x))/h#
So given:
#g(x) = 1/(x-2)#
its average rate of change over
#(1/(color(blue)((x+h))-2)-1/(color(blue)(x)-2))/h = ((color(red)(cancel(color(black)(x)))-color(brown)(cancel(color(black)(2))))-(color(red)(cancel(color(black)(x)))+h-color(brown)(cancel(color(black)(2)))))/(h(x-2)(x+h-2))#
#color(white)((1/(((x+h))-2)-1/((x)-2))/h) = (-color(red)(cancel(color(black)(h))))/(color(red)(cancel(color(black)(h)))(x-2)(x+h-2))#
#color(white)((1/(((x+h))-2)-1/((x)-2))/h) = -1/((x-2)(x-2+h))#
In particular, we find:
#lim_(h->0) (g(x+h)-g(x))/h = -1/(x-2)^2#
This is the slope of